Unfortunately this answer is incorrect and is a common misunderstanding of big-O. Selection sort is typically stated to be O(n^2), which is true, but the way that O works is that it's a bound from above. It means "the set of functions that asympotically are guaranteed to not take more time than".
Selection sort is also O(n^3) and O(n!) and O(n^2 log log log log log n). The correct operator for saying that f(x) is O(n) is the set exists operator ∊. O(n^2) is a set of functions that is a strict subset of the set of functions that are in O(n^3).
Selection sort is only big-theta(n^2), which indicates both the bounds are tight (eg for Selection sort it is true that it's asymptotic behavior is both f(n) <= an^2 + m AND f(n) >= bn^2 + n.
Selection sort is also O(n^3) and O(n!) and O(n^2 log log log log log n). The correct operator for saying that f(x) is O(n) is the set exists operator ∊. O(n^2) is a set of functions that is a strict subset of the set of functions that are in O(n^3).
Selection sort is only big-theta(n^2), which indicates both the bounds are tight (eg for Selection sort it is true that it's asymptotic behavior is both f(n) <= an^2 + m AND f(n) >= bn^2 + n.